Problem 51 - Algebra 2

Question

Assume that \(\log 4 \approx 0.6021, \log 5 \approx 0.6990,\) and \(\log 6 \approx 0.7782 .\) Use the properties of logarithms to evaluate each expression. Do not use your calculator. \(\log \frac{1}{25}\)

Step-by-Step Solution

Answer

The value of \(\log \frac{1}{25}\) approximates to -1.3980.

Step 1: Convert to Logarithmic Form with Base

Convert the logarithm \(\log \frac{1}{25}\) into a form where you can apply the properties of logs. Since \(\frac{1}{25} = \frac{1}{5^2}\), let's rewrite \(\log \frac{1}{25}\) as \(\log(5^{-2})\).

Step 2: Apply Power Rule of Logarithms

The power rule of logarithms states that \(log_x (a^n) = n \cdot log_x a\). Hence, we rewrite \(\log(5^{-2})\) as \(-2 \log 5\).

Step 3: Use Given Logarithmic Values

Replace \(\log 5\) with its approximate value of 0.6990 given in the problem. Thus, \(-2 \log 5\) becomes \(-2 \times 0.6990\).

Step 4: Perform Multiplication

Multiply -2 by 0.699 to get the result of \(-2 \times 0.6990 = -1.3980\).

Key Concepts

Properties of Logarithms

Logarithms possess several useful properties that make them powerful tools for simplifying expressions and solving equations. These properties are rules that govern how logarithms can be manipulated.

• Product Rule: This rule states that the logarithm of a product is the sum of the logarithms of the factors: \(\log_b(xy) = \log_b x + \log_b y\).
• Quotient Rule: According to this rule, the logarithm of a quotient is the difference of the logarithms: \(\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y\).
• Power Rule: This rule indicates that the logarithm of a power can be rewritten as a multiple of the logarithm: \(\log_b(x^n) = n \cdot \log_b x\).

These properties allow us to break down complex logarithmic expressions into simpler components.
For example, knowing \(\log 5\) allows us to evaluate expressions involving \(5^2\) by employing the power rule.

Logarithmic Expressions

Logarithmic expressions are algebraic forms that involve logarithms. Understanding how to manipulate these expressions is critical for efficiently solving math problems.

A logarithmic expression can represent various combinations of base, exponent, and logarithmic values. For instance,

• The expression \(\log\frac{1}{25}\) involves recognizing a fraction and applying logarithmic rules to simplify it.
• Recognizing equivalent forms, such as \(\log(5^{-2})\), provides an opportunity to use the power rule.

Different problems might need different properties. Recognizing how to restructure expressions using known values (like \(\log 4 \approx 0.6021\) or \(\log 5 \approx 0.6990\)) streamlines solving processes.
Sophisticated handling of logarithmic expressions allows for translating complex values into simple arithmetic calculations.

Power Rule of Logarithms

The power rule of logarithms is a particularly helpful tool in simplifying and resolving logarithmic expressions. This rule indicates that the logarithm of an exponentiated expression can be rewritten as a product:\[\log_b(a^n) = n \cdot \log_b a\].

• When applied, this rule allows you to "bring down" the exponent in front of the logarithm, turning it into a straightforward multiplication problem.
• In our example, converting \(\log(5^{-2})\) to \(-2 \cdot \log 5\) exploits this rule to achieve a simpler calculation path.

Such transformations can significantly reduce the complexity of a problem, especially when direct computation isn't feasible (like without using a calculator).
Mastering the power rule gives an edge in efficiently transforming and simplifying expressions, paving the way for accurate and easy computations.